Normality and the Law of Equivalence

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. I take two beakers. In one, hydrochloric acid. In the other, sulphuric acid. Same molarity, same volume, so the same number of moles of acid in both, okay? Now I neutralise each one with the same sodium hydroxide solution. And the sulphuric acid needs double the base! Same moles, but double the base? Wait, how is that possible? By the end of this video, you will explain it in one line. And look, almost every titration question in JEE runs on today's idea: normality, and the law of equivalence. So, let's begin.

0:44Normality: count equivalents, not moles

So, first question. In our puzzle, the moles of acid were equal, and still one acid needed double the base. So counting moles alone does not tell the full story, right? Then what should we count? Yes, equivalents! You met them in the last lesson. The number of equivalents per litre of solution is called normality. Capital N equals the equivalents divided by V, the volume of the solution in litres. When a capital N is written after a number, we read it as normal. Clear? Now, how do we get normality from molarity? Just multiply. N equals capital M, the molarity, times the n factor. And remember, for an acid, the n factor is the number of replaceable H plus ions per molecule that are actually used, okay? So, hydrochloric acid has n factor one, and sulphuric acid has n factor two. See what this means. Same molarity, but sulphuric acid has double the normality! Keep this in mind. We will come back to our puzzle in a minute. Now a topper's trick. In the lab, volumes come in millilitres, right? So why convert to litres at all? Milliequivalents equals N times V in millilitres, directly. On screen we write meq for short, and we will use it in every example today.

2:10The law of equivalence

Alright, now the most important law of this chapter. When an acid reacts with a base, do they always react in equal moles? Not moles. Equivalents! Substances always react in equal numbers of equivalents. This is called the law of equivalence. Moles may or may not match, but equivalents always match. That's it. That is the whole law. So in a titration, the two solutions react in equal equivalents. And normality times volume gives equivalents, right? So, N one V one equals N two V two. The little one belongs to the first solution, and the little two to the second, okay? One line, and every titration question opens up. Now, our puzzle. Same molarity. But sulphuric acid has n factor two, so its normality is two times that of hydrochloric acid. Same volume, so double the equivalents. And equivalents must match, so it needs double the base! Same moles, double the equivalents. Puzzle solved. Got it?

3:17Mixing solutions

Now, one more situation. What if I pour two solutions together? What is the normality of the mixture? Simple. Add up the milliequivalents of all the solutions, and divide by the total volume. N mix equals the sum of N i V i, divided by the sum of V i. The big sigma you see on screen just means add them all up, and the little i stands for each solution, okay? If both solutions are acids, nothing cancels, so their milliequivalents simply add. Easy, right? But if one is an acid and the other is a base? They cancel each other. So the milliequivalents subtract, and only the excess is left, spread over the total volume. In JEE, this subtraction is exactly the step students forget. Don't you forget it!

4:05Example 1: volume of NaOH to neutralise 0.2 M H2SO4

Okay, example one. What volume of zero point one normal sodium hydroxide neutralises twenty five millilitres of zero point two molar sulphuric acid? Pause the video, and try it first. Done? Okay, let's check. Many students write zero point two times twenty five equals zero point one times V, and get fifty millilitres. Fifty, right? No, no, careful! Look at the question again. The acid is given in molarity, not in normality. Sulphuric acid has n factor two. So its normality is zero point two times two: zero point four normal. Clear? Now the law of equivalence. N one V one equals N two V two. So, zero point four times twenty five equals zero point one times V, okay? Zero point four times twenty five is ten milliequivalents of acid. So V is ten divided by zero point one: one hundred millilitres. Not fifty, one hundred. Double! Exactly like our puzzle: the n factor of two doubles the base. Just as we expected.

5:17Example 2: normality of acid and acid-base mixtures

Now a harder one, JEE level. Read it with me. One hundred millilitres of zero point one normal hydrochloric acid is mixed with one hundred millilitres of zero point two normal sulphuric acid. Find the normality of the mixture. Then, instead, fifty millilitres of zero point two normal hydrochloric acid is mixed with fifty millilitres of zero point one normal sodium hydroxide. What is the normality now? Pause and try. Hint: count milliequivalents first. Done? Okay, let's check. Part one. Both are acids, so the milliequivalents add. One hundred times zero point one is ten. One hundred times zero point two is twenty. Total, thirty milliequivalents. Now divide by the volume. But which volume? The total volume, yes! One hundred plus one hundred: two hundred millilitres, not one hundred. Thirty divided by two hundred: zero point one five normal, and the mixture is acidic. Got it? Part two. Now an acid meets a base. The acid: fifty times zero point two, ten milliequivalents. The base: fifty times zero point one, five milliequivalents. So, ten plus five, fifteen milliequivalents in total, right? No, no! This is the JEE trap. The acid and the base cancel each other. So we subtract. Ten minus five: five milliequivalents of acid are left over, in a total of one hundred millilitres. Five divided by one hundred: zero point zero five normal, and the solution is still acidic. See? Acid with acid, add. Acid with base, subtract. Really simple.

7:02Quick revision: N = M x n-factor and N1V1 = N2V2

So, quick revision. Normality is molarity times the n factor, and normality times millilitres gives milliequivalents. Substances always react in equal equivalents, so N one V one equals N two V two. And in mixtures, acids add their milliequivalents, but an acid and a base subtract. And our puzzle? Same moles, but double the equivalents, so double the base. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!